If $|\alpha|=\nu$, then $\lambda_{1}=\lambda_{2}=-\alpha$ and the corresponding generalized eigenspace is spanned by
$$
z^{1}=\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right], \quad z^{2}=\left[\begin{array}{l}
0 \
1
\end{array}\right]
$$
where $z^{1}$ is a real eigenvector and $z^{2}$ is a generalized eigenvector of second order. By (34) the corresponding (generalized) real eigenmotions are
$$
x^{1}(t)=e^{-\alpha t}\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right], \quad x^{2}(t)=e^{-\alpha t}\left[\begin{array}{l}
0 \
1
\end{array}\right]+t e^{-\alpha t}\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right]
$$
If $|\alpha|<v$ (hence $z^{2}=\overline{z^{1}}$ ) then $\lambda_{1,2}=-\alpha \pm x \sqrt{\nu^{2}-\alpha^{2}}$ and the corresponding real modes are by $(44),(45)$ and $(50)$
$$
x^{1}(t)=e^{-\alpha t}\left(\cos \left(\sqrt{u^{2}-\alpha^{2}} t\right)\left[\begin{array}{c}
1 \
-\alpha
\end{array}\right]-\sin \left(\sqrt{u^{2}-\alpha^{2}} t\right)[\sqrt{0}]\right)
$$
MTH2040 COURSE NOTES ：
is called the Fibonacci sequence and plays a role in various fields of mathematics. The eigenvalues of the matrix in (59) are $\lambda_{1,2}=(1 \pm \sqrt{5}) / 2$ and the corresponding real modes are
$$
z^{1}(t)=\left(\frac{1+\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c}
(1+\sqrt{5}) / 2 \
1
\end{array}\right], \quad z^{2}(t)=\left(\frac{1-\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c}
(1-\sqrt{5}) / 2 \
1
\end{array}\right]
$$
The solution of the initial value problem is of the form $x(t)=\alpha_{1} z^{1}(t)+\alpha_{2} z^{2}(t)$ for some $\left(\alpha_{1}, \alpha_{2}\right) \in \mathbb{R}^{2}$. Since $x(0)=[1,0]^{\top}$ we obtain $\alpha_{1}=1 / \sqrt{5}, \alpha_{2}=-1 / \sqrt{5}$ and hence the following analytic expression for the Fibonacci sequence
$$
\xi(t)=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{t+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{t+1}\right]
$$